Display individual elements from grouped rows on same row

I want to both group some rows together but return 1 column of the individual rows from the group onto the same line as well, probably easiest to show with an example….

Original Data

rDate      | track | horse | odds
-----------------------------
01/06/2021 | Ascot | Jim   | 3
01/06/2021 | Ascot | Dave  | 3
01/06/2021 | Ascot | Rex   | 15
01/06/2021 | Epsom | Lee   | 2
01/06/2021 | Epsom | Ben   | 3
01/06/2021 | Epsom | Gary  | 4

Grouped by rDate, track, AVG(odds)

Select rDate, track, AVG(odds)
FROM Results
GROUP BY rDate, track

rDate      | track | AvgOdds
-----------------------------
01/06/2021 | Ascot | 7
01/06/2021 | Epsom | 3

Desired output, grouped with average odds but with individual odds added to grouped row as well:

rDate      | track | AvgOdds | odds1 | odds2 | odds3
--------------------------------------------------
01/06/2021 | Ascot |    7    |   3   |   3   |   15
01/06/2021 | Epsom |    3    |   2   |   3   |   4

The number of rows in each group varies from 4 to 8, but a solution using a fixed number of rows would be acceptable, I can work around it.

Answer

I would suggest conditional aggregation phrased like this:

SELECT rDate, track, AVG(odds) as avgodds,
       MAX(CASE WHEN seqnum = 1 THEN odds END) as odds1,
       MAX(CASE WHEN seqnum = 2 THEN odds END) as odds2,
       MAX(CASE WHEN seqnum = 3 THEN odds END) as odds3
FROM (SELECT r.*,
             ROW_NUMBER() OVER (PARTITION BY rDate, Track ORDER BY odds) as seqnum
      FROM Results r
     ) r
GROUP BY rDate, track;

Note: If you want to extend this to 8 columns for odds, just follow the pattern in the SELECT.